Class 7

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 8 Congruency of Triangles Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Congruency of Triangles Exercise 1

Question 1.
Decide whether the SSS congruence is true with the following figures. Give reasons

Solution:
i) From the figures,
BE = CA
EN = AR
BN = CR
∴ S.S.S congruency is true.

ii) From the figure,
AL = SD = 4 cm
LD = LD = common side
AD ≠ LS (3 cm ≠ 2.5 cm)
∴ S.S.S congruency is not ture.

Question 2.
For the following congruent triangles, and the pairs of corresponding angles.

Solution:
i) ∠P = ∠R
∠T = ∠S
∠TQP = ∠SQR

ii) ∠P=∠S
∠Q = ∠R
∠POQ = ∠SOR

Question 3.
In the adjacent figure, choose the correct answer!
(i) ΔPQR ≅ ΔPQS
(ii) ΔPQR ≅ ΔQPS
(iii) ΔPQR ≅ ΔSQP
(iv) ΔPQR ≅ ΔSPQ

Solution:
(ii) ΔPQR ≅ ΔQPS is correct.

Question 4.
In the figure given below, AB = DC and AC = DB. Is ΔABC ≅ ΔDCB.

Solution:
Given that AB = DC
AC = DB
Also from the figure ∠B = ∠C
∴ ΔABC ≅ ΔDCB

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 3

Question 1.
A length of a bacteria enlarged 50,000 times attains a length of 5 cm. What is the actual length of the bacteria? If the length is enlarged 20,000 times only, what would be its enlarged length?
Solution:
After enlarging 50,000 times the length is 5 cm.
Without enlarging (1 – times) the length Is x cm say
∴ 50,000 : 5 :: 1 : x
By rule of properties
50,000 × x = 5 × 1
x = \(\frac{5}{50,000}=\frac{1}{10,000}\) = 0.0001 cm

Question 2.
Observe the following tables and fmd if x is directly proportional.
Solution:
(i)


In all the cases \(\frac { x }{ y }\) is constant.
∴ x ∝ y (x is directly proportional toy)

(ii)

Here the change/increase is not uniform.
(i.e.,) \(\frac { x }{ y }\) is not same in all cases.
∴ x is not directly proportional to y.

(iii)

Here the change/increase is not uniform.
(i.e.,) \(\frac { x }{ y }\) is not same in all cases.
∴ x is not directly proportional to y.

Question 3.
Sushma has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Solution:
Scale of the map is 1 cm = 18 km
Distance covered = 72 km
Let it be xcm on the map
then 1 : 18 :: x : 72
By rule of proportion
Product of extremes Product oÍ means
18x = 72 × 1
x = \(\frac{72}{18}\) = 4cm

Question 4.
On a Grid paper, draw five squares of different sizes. Write the following information in a tabular form.

Find whether the length of a side is in direct proportion to:
(i) the perimeter of the square.
(ii) the area of the square.
Solution:

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 4

Question 1.
In ΔABC, name all the interior and exterior angles of the triangle.

Solution:
Interior angles
∠ABC, ∠BAC, ∠CAB
Exterior angles
∠ACZ, ∠BAY and ∠CBX

Question 2.
For ΔABC, find the measure of ∠ACD.

Solution:
In ΔABC,
∠ACD = ∠BAC + ∠ABC (exterior angle property)
=66° + 45° = 111°.

Question 3.
Find the measure of angles x and y.

Solution:
In the figure,
x° + 65° = 180° (lInear pair of an1es)
∴ x°= 180°- 65°= 115°
Also y° + 30° = 65° (exterior angle property)
y° = 65° – 30° = 35°

Question 4.
In the following figures, fmd the values of x and y.


Solution:
In ΔABC,
107° = x° + 57° (exterIor angle property)
∴ x° = 107° – 57° =50°
In ΔADC,
∠CAD + ∠ADC + ∠ACD = 180° (angle – sum property)
40° + 107°+ x°= 180°
∴ x° = 180° – 147° = 33°
Also in ΔABC,
∠A +∠B +∠C = 180°
40° + ∠B + (33° + 65°) = 180°
∠B + 138° = 180°
B = 180° – 138° = 42°
Now y° = ∠A + ∠B for ΔABC (exterior angle property)
=40° + 42° = 82°

Question 5.
In the figure ∠BAD = 3 ∠DBA, find ∠CDB. ∠DBC and ∠ABC.

Solution:
In ΔBCD,
∠DBC + ∠BCD = ∠BDA (exterior angle property)
∠DBC + 65° -104°
∴ ∠DBC = 104° – 65° = 39°
Also ∠BDA + ∠BDC = 180° (linear pair of angles)
104° + ∠BDC = 180°
∠CDB or ∠BDC = 180° – 104° = 76°
Now in ΔABD,
∠BAD + ∠ADB + ∠DHA = 180°
3∠DBA + ∠DBA + 104° = 180° (given)
4∠DBA + 104° – 180°
∴ 4∠DBA = 180° – 104°= 76°
∴∠DBA = \(\frac{76^{\circ}}{4}\) = 19°
Now ∠ABC = ∠DBA + ∠DBC = 19° + 39° = 58°

Question 6.
Find the values of x and y in the following figures.




Solution:
i) In the figure the angles are
70°, x°, x° (angles in an Isosceles triangle)
Also 70° + x° + x° = 180° (angle – sum property)
2x° + 70°= 180°
2x° = 180° – 70° [∵ y° = x°]
2x° = 110°
x°= \(\frac{110^{\circ}}{2}\) = 55°

ii) From the figure,
the Interior opposite angles of x are 50°, 500 (angles in an isosceles triangle )
50° + 50° = x° (exterior angle is equal to sum of the Interior opposite angles)
∴ x = 100°

iii) From the figure,
y° = 30° (vertically opposite angles)
ALso y = x° – 30° (equal angles of an Isosceles triangles)
∴ x° = 30° and y = 30°

iv) From the figure,
a + 110° = 180° (linear pair of angles)
∴ a = 180° – 110°= 70
Also y° = a° = 70° (equal angles of an isosceles triangle)
x + y + a = 180 (sum of interior angles)
x + 70 + 70 = 180
x + 140 = 180
x = 180 – 40 = 40°, ∴ x = 40°

v) From the figure,
30° + y° – 180°
y° = 180° – 30° = 150°
Also x° + a° = y° (exterior angle property)
x° + 90° = 150°
x° = 150° – 90° = 60°

vi) From the figure,
b = 80° (vertically opposite angles)
Also x° = a° (equal angles of an isosceles triangle.)
° 80° + x° + a° = 180
80 + x° + x° = 180
2x° = 180° – 80° =100°
2x° = 100°
∴ x° = \(\frac{100}{2}\) = 50°
Now y = x° b° (exterior angle property)
50° + 80° = 130°

Question 7.
One of the exterior angles of a triangle is 125° and the interior opposite angles are in the
ratio 2 :3. Find the angles of the triangle.
Solution:
Ratio of the interior opposite angles = 2 : 3
Sum of the terms of the ratio = 2 + 3 = 5
Sum of the interior angles exterior angle = 125°
∴ First angle = \(\frac{2}{5}\) x 125° = 500
Second angle = \(\frac{3}{5}\) x 125° = 75°

Question 8.
The exterior ∠PRS of ∆PQR is 105°. If Q = 70°. find ∠P. Is ∠PRS > ∠P?
Solution:

∠P+ ∠Q = ∠PRS (exterior angle is equal to sum of the interior opposite angles)
∠P + 70° = 105°
∠P = 105° – 70° = 35°
Now ∠PRS > ∠P.

Question 9.
If an exterior angle of a triangle is 130° and one of the interior opposite angle is 6. Find
the other interior opposite angle.
Solution:
Let the other interi6r opposite angle be = x°
Give interior opposite angle be = 60°
Now sum of the interior opposite angle = exterior angLe
x° + 60° = 130°
x° = 130° – 60° = 70°
∴ The other interior opposite angle = 70°

Question 10.
One of the exterior angle ofa triangle is 105° and the interior opposite angles are in the ratio 2 : 5. Find the angles of the triangle.
Solution:
Ratio of interior opposite angles = 2 : 5
Sum of the terms of the ratio = 2 + 5 = 7
Sum of the angles = 105°
∴ 1st angle = \(\frac { 2 }{ 7 }\) x 105° = 30°
2nd angIe = \(\frac { 5 }{ 7 }\) x 105° = 75°
3rd angle = 180° – (30° + 75°) [∵ angle – sum property]
= 180° – (105°) = 75°

Question 11.
In the figure find the values of x andy.
Solution:

From the figure,
∠y = 50° + 30° = 80° (exterior angle property)
∠x = y° + 55° (exterior angle property)
= 80° + 55°
= 135°

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 6 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 6

Question 1.
How long will it take for a sum of ₹ 12600 invested at 9% per annum beocme to ₹ 15624?
Solution:
HereA = ₹ 15,624 R = 9% T = ?
P = ₹ 12,600
∴ I = A – P = 15,624 – 12,600 = ₹ 3024
Also I = \(\frac{P R T}{100}\)
∴ 3024 = \(\frac{12,600 \times 9 \times \mathrm{T}}{100}\)

Question 2.
At what rate a sum doubles itself in 8 year 4 months?
Solution:
Given that A = double the sum
Let the principle be P
then A = 2P
T =8 years 4 months = 8y \(\frac { 4 }{ 12 }\) y = 8\(\frac { 1 }{ 3 }\) years = \(\frac { 25 }{ 3 }\)
R = R% say
We know that
I = \(\frac{\text { PTR }}{100}\)
Here I = A – P
= 2P – P = P
∴ P = \(\frac{P \cdot \frac{25}{3} \times R}{100}\)
∴ R = 100 x \(\frac{3}{25}\) = 12%

Question 3.
A child friendly bank announces a savings scheme for school children. They will give kiddy banks to children. Children have to keep their savings in it and the bank collects all the money once in a year. To encourage children savings, they give 6% interest if the amount exceeds by ₹ 10000. and other wise 5%. Find the interest received by a school if they deposit is ₹ 9000 for one year.
Solution:
Money deposited = ₹ 9000
Interest applicable 5% on ₹ 9000
= 5 × \(\frac{9000}{100}\) = ₹ 450

Question 4.
A sum of money invested at 8% per annum for simple interest amounts to ₹ 12122 in 2 years. What will it amounts to in 2 year 8 months at 9% rate of interest?
Solution:
First part
Pirnciple = P say
R=8%
T = 2 years
A = ₹ 12,122

Second part
P = ₹10,450
R = 9%
T = 2 years 8 months, A = ?

Question 5.
In 4 years, ₹ 6500 amounts to ₹ 8840 at a certain rate of interest. In what time will ₹ 1600 amounts to 1816 at the same rate?
Solution:
First part
T = 4 years
P = ₹ 6500
A = ₹ 8840
R = R% = ?
I = A – P = 8840 – 6500 = ₹ 2340
But I = \(\frac{P T R}{100}\)
∴ 2340 = \(\frac{6500 \times 4 \times R}{100}\)
∴ R = \(\frac{2340 \times 10 \theta}{6500 \times 4}=\frac{36}{4}\) = 9%

Second part
P = ₹ 1600
A = ₹ 1816
T = ?
R = 9%
I = A – P = 1816 – 1600 = ₹ 216
But I = \(\frac{P T R}{100}\)
∴ 2340 = \(\frac{1600 \times T \times 9}{100}\)
∴ R = \(\frac{216 \times 100 \theta}{1600 \times 9}=\frac{3}{2}\) = 1 \(\frac { 1 }{ 2 }\) years.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 5 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 5

Question 1.
A shopkeeper bought a suit case for ₹ 480 and sold it for ₹ 540. Find his gain percent?
Solution:
C.P of the suitcase = ₹ 480
SP of the suitcase = ₹ 540
∴ gain=S,P-C.P = 540 – 480 = ₹ 60
∴ gain percent = \(\frac { 60 }{ 450 }\) × 100% = 12.5%

Question 2.
Ajay bought a TV for ₹ 15000 and sold it for ₹141 00. Find the loss percent?
Solution:
C.P of the TV = ₹ 15,000
S.P of the TV = ₹ 14100
As S.P < C.P
Loss = C.P – S.P= 15,000 – 14,100 = ₹ 900
∴ Loss percent = \(\frac { 900 }{ 15000 }\) × 100 = 6%

Question 3.
Ramu sold a plot of land for ₹ 2,40,000 gaining 20%. For how much did he purchase the plot?
Solution:
Let the cost price of the land = ₹ x
gain percent = 20% of the C.P
∴ S.P of the land = 120% of C.P
By problem, 120% of x = 2,40,000
\(\frac { 120. x }{ 100 }\) = 2,40,000
x = 2.40000 × \(\frac {100 }{ 120 }\)
= ₹ 2,00,000

(OR)

Let the C.P of the land = ₹ x
gain percent = 20% of C.P = 20% of x = \(\frac{20 \mathrm{x}}{100}\)
∴ S.P of the land = C.P + gain
= x + \(\frac{20 \mathrm{x}}{100}\)
= \(\frac{100 x+20 x}{100}=\frac{120 x}{100}\)

By problem, \(\frac{20 \mathrm{x}}{100}\) = 2,40.000
∴ x = 24000 x \(\frac { 100 }{ 120 }\)
= ₹ 2,00,000

Question 4.
On selling a mobile for ₹ 750, a shop keeper looses 10%. For what amount should he sell it to gain 5%?
Solution:
Let the C.P of the mobile be x
Losspercent = 10% of C.P = 10% of x = \(\frac{10 \mathrm{x}}{100}\)
∴ Loss = \(\frac{10 x}{100}\)
∴ S.P = C.P – Loss
= \(x – \frac{10 x}{100}=\frac{100 x-10 x}{100}=\frac{90 x}{100}\)

By problem, \(\frac{90 \mathrm{x}}{100}\) = 750
x = 750 × \(\frac { 100 }{ 90 }\) = ₹ \(\frac { 2500 }{ 3 }\)
Now to gain by 5% the S.P = C.P + 5% of C.P

Question 5.
A farmer sold 2 bullocks for ₹ 24000 each. On one bullock he gained 25% and on the other he lost 20%. Find his total profit or loss percent?
Solution:
Let the C.P of first bullock be x
gain = 25% of x = 25 × \(\frac{x}{100}=\frac{25 x}{100}\)
S.P = C.P + gain = x + \(\frac{25 x}{100}=\frac{100 x+25 x}{100}=\frac{125 x}{100}\)
Selling price of first bullock 24,000 (given)
\(\frac{125 x}{100}\) = 24000
x = \(\frac{24000}{125}\) × 100
∴ C.P of first bullock = ₹ 19.200
Let the C.P of second bullock be y
Loss percent = 20%

∴Loss = \(\frac{20 \times \mathrm{y}}{100}=\frac{20 \mathrm{y}}{100}\)

S.P = C.P – loss
= \(y-\frac{20 y}{100}=\frac{100 y-20 y}{100}=\frac{80 y}{100}\)

But S.P = ₹ 24,000 (given)
\(\frac{80 \mathrm{y}}{100}\) = 24,000
∴ y = 24000 x \(\frac{100}{80}\) = 30,000

∴ C.P of second bullock = ₹ 30,000
C.P of 2 bullocks = ₹19,200 + ₹30,000
= ₹ 49,200
S.P of 2 bullocks ₹24,000 × 2 = ₹48,000
Loss = C.P – S.P = 49,200 – 48,000 = ₹1200
Loss % = \(\frac{\text { loss }}{\text { CP }}\) × 100
= \(\frac{1200}{49200}\) × 100 = 2.4
∴ Loss% = 2.4%

Question 6.
Sravya bought a watch for ₹480. She sold it to Ridhi at a gain of 6\(\frac { 1 }{ 4 }\) %. Ridhi sold it to Divya at a gain of 10%. How much did Divya pay for it?
Solution:
CF of the watch = ₹ 480
Gain% = \(6 \frac{1}{4} \%=\frac{25}{4} \%\)
Gain = 480 × \(\frac{25}{4}\) x \(\frac{1}{100}\) = 30
C.P of the watch for Ridhi = 480 + 30 = ₹ 510
Gain% = 10%
∴ Gain= 51% × \(\frac{10}{100}\) = 51
S.P of the watch for Ridhi = 510 + 51 = ₹561
∴ Amount paid for Divya for the watch = ₹561

Question 7.
The marked price of a book is ₹225.Thc publisher allows a discount of ₹10% on it. Find the selling price of it?
Solution:
The marked price of the book = ₹ 225
Discount 10%
∴ Discount = 225 × \(\frac{10}{100}\)= ₹ 22.5
S.P = M.P – discount
= 225 – 22.5 = ₹202.5
∴ Selling price of the book = ₹ 202.5

Question 8.
A carpenter allows 15% discount on his goods. Find the marked price of a chair which is sold by him for ₹680?
Solution:
Let the marked price of chair be ₹x
Discount = 15% = x × \(\frac{15}{100}=\frac{15 x}{100}\)
S.P = M.P – discount
\(x-\frac{15 x}{100}=\frac{100 x-15 x}{100}=\frac{85 x}{100}\)
S.P = ₹ 680 (given)
∴ \(\frac{85 \mathrm{x}}{100}\) = 680
x = \(\frac{680 \times 100}{85}\) = 800
Marked price of the chair = ₹ 800

Question 9.
A dealer allows a discount of ₹ 10% and still gains by 10%. What should be the marked price if the cost price is ₹900?
Solution:
Given that C.P = 900
There is a gain of 10% on C.P
∴ Gain = 10% of ₹ 900
= \(\frac{10 \times 900}{100}\) = ₹ 90
∴ S.P = C.P . gain
= 900 + 90
S.P = ₹990 …………….. (1)
Let the marked price = ₹ x
Discount on it = 10% on M.P = \(\frac{10 \mathrm{x}}{100}\)
∴ Selling Price = MP – Discount
= x – \(\frac{10 \mathrm{x}}{100}\)
= \(\frac{100 x-10 x}{100}\)
∴ S.P = \(\frac{90 \mathrm{x}}{100}\) …………….(2)
From (1) & (2) \(\frac{90 \mathrm{x}}{100}\) = 990
∴ x = 990 x \(\frac{100}{90}\)
∴ M.P = x = ₹1100

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 6 Ratio – Applications Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 6th Lesson Ratio – Applications Exercise 4

Question 1.
In a school X, 48 students appeared for 10th class exam out of which 36 students passed. In another school Y, 30 students appeared and 24 students passed. If the District Educational Officer wants to give an award on the basis of pass percentage. To which school will he give the award?
Solution:
From the first school:
Number of students appeared for the exam = 48
Number of students passed = 36
as a percentage = \(\frac { 36 }{ 48 }\) × 100 = 75
Pass percentage = 75%

From the second school:
Number of students appeared = 30
Number of students passed = 24
Pass percentage = \(\frac { 24 }{ 30 }\) × 100 = 80%
As 80% > 75%
D.E.O should give award to the 2nd school.

Question 2.
Last year the cost of 1000 articles was ₹ 5000. This year it goes down to ₹ 4000. What is the percentage of decrease in price?
Solution:
Last year
Cost of 1000 articles = ₹ 5000
∴ Cost of one article = \(\frac { 5000 }{ 1000 }\) = ₹ 5
This year
Cost of 1000 articles = ₹ 4000
∴ Cost of one article = \(\frac { 4000 }{ 1000 }\) = ₹ 4
Decrease in price = ₹ 5 – ₹ 4 = ₹ 1
Decrease In percentage = \(\frac { 1 }{ 5 }\) × 100% = 20%

Question 3.
Sri Jyothi has a basket full of bananas, oranges and mangoes. If 50% are bananas, 15% are oranges, then what percent are mangoes?
Solution:
Percent of bananas = 50%
Percent of oranges = 15%
∴ Percent of mangoes = 100% – (50% + 15%)
= 100% – 65% = 35%

Question 4.
64% + 20% + …..? = 100%
Solution:
(64% + 20%) +…………… = 100%
∴ ?=100% – 84% = 16%

Question 5.
On a rainy day, out of 150 students in a school 25 were absent. Find the percentage of students absent from the school? What percentage of students are present?
Solution:
Total students = 150
Number of students absent = 25
∴ Number of students present = 150 – 25 = 125

Question 6.
Out of 12000 voters in a constituency, 60% voted. Find hte number of people voted in the
constituency?
Solution:
Percentage of people who voted = 60% of I 2,000
∴ Number of people voted = \(\frac{60 \times 12000}{100}\) = 7,200

Question 7.
A local cricket team played 20 matches in one season. If it won 25% of them and lost rest. How many matches did it loose?
Solution:
Total number of matches played = 20
Matches won = 25%
∴ Matches lost = (100 –  25)% of total played
= 75% of 20
\(\frac{75 \times 20}{100}\) = 15

Question 8.
In every gram of gold, a goldsmith mixes 0.25 grams of silver and 0.05 grams of copper. What is the percentage of gold, silver and copper in every gram of gold?
Solution:
In 1 gm
The quantity of silver = 0.25 gm
The quantity of copper = 0.05 gm
∴ The quantity of gold = 1 gm – (0.25 . 0.05)
= 1 – 0.3 = 0.7gm
∴ Percentage of gold = \(\frac { 0.7 }{ 1 }\) × 100 = 70%
∴ Percentage of silver = \(\frac { 0.25 }{ 1 }\) × 100 = 25%
∴ Percentage of copper = \(\frac { 0.05 }{ 2 }\) × 100 = 5%

Question 9.
40% of a number is 800 then find the number?
Solution:
Let the number be = x
40% of x = 800
(i.e.,) \(\frac{40 \times x}{100}\) = 800
x = \(\frac{800 \times 100}{40}\)
x = 2000
∴ The required number = 2000

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 1

Question 1.
Solve the following.
(i) 2 + \(\frac { 3 }{ 4 }\)
(ii) \(\frac{7}{9}+\frac{1}{3}\)
(iii) 1 – \(\frac{4}{7}\)
(iv) \(2 \frac{2}{3}+\frac{1}{2}\)
(v) \(\frac{5}{8}-\frac{1}{6}\)
(vi) \(2 \frac{2}{3}+3 \frac{1}{2}\)
Solution:

Question 2.
Arrange the following in ascending order.
(i) \(\frac{5}{8}, \frac{5}{6}, \frac{1}{2}\)
(ii) \(\frac{2}{5}, \frac{1}{3}, \frac{3}{10}\)
Solution:

Question 3.
Check whether in this square the sum of the numbers in each row and in each column and along the diagonals is the same.

Solution:

Question 4.
A rectangular sheet of paper is 5\(\frac{2}{3}\) cm long and 3\(\frac{1}{5}\) cm wide. Find its perimeter.
Solution:
Length of the rectangular sheet = 5\(\frac{2}{3}\) cm
Breadth/width of the rectangular sheet = 5\(\frac{2}{3}\) cm
Perimeter = 2 x (length + breadth)

Question 5.
The recipe requires 3\(\frac{1}{4}\) cups of flour. Radha has 1\( \frac{3}{8}\) cups of flour. How many more cups of flour does she need?
Solution:
Flour required for the recipe = 3\(\frac{1}{4}\) cups
Flour with Radha = 1\(\frac{3}{8}\) cups
More cups of flour required = \(3 \frac{1}{4}-1 \frac{3}{8}\)
= \(\frac{3 \times 4+1}{4}-\frac{1 \times 8+3}{8}\)

Question 6.
Abdul is preparing for his final exam. He has completed \(\frac{5}{12}\) part of his course content. Find out how much course content is left?
Solution:
Take content as 1 (i.e., full) Course completed = \(\frac{5}{12}\)
Course yet to be completed = 1 – \(\frac{5}{12}\)
= \(\frac{12 \times 1-5}{12}\)
\(\frac{12-5}{12}=\frac{7}{12}\)

Question 7.
Find the perimeters of(i) ΔABE (ii) the rectangle BCDE in this figure. Which figure has greater perimetre and by how much?

Solution:
i) Perimeter of ΔABE = AB + BE + AE

ii) Perimeter of BCDE = 2(BE + BC)

As \(\frac{116}{15}<\frac{153}{15}\), we conclude that the perimetre of ΔABE > Perimeter of BCDE

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 3

Question 1.
Find the value of the unknown ‘x’ in the following triangles.

Solution:
i) In ΔABC

∠A + ∠B + ∠C = 180° (angle sum property
x° + 50° + 60° = 180°
x° + 110° = 180°
x° = 180° – 110°

ii) In ΔPQR,

∠P + ∠Q + ∠R = 180° (angle – sum property)
90° + 30° + x° = 180°
120° + x° – 180°
x° = 180° – 120°
∴ x° = 60°

iii) In ΔXYZ,

∠X + ∠Y + ∠Z = 180° (angle – sum property)
30°+110° + x° =180°
140° + x° = 180°
x° = 180° – 140°
∴ x° = 40°

Question 2.
Find the values of the unknowns ‘x ‘and ‘y ‘in the following diagrams.
Solution:
i) In ΔPQR,

x° + 50° = 120° (exterior angle property)
x°= 120°- 50°
x°= 70°
Also
∠P + ∠Q +∠R = 180° (angle – sum property)
70° + 50° + y° = 180°
120° + y° = 180°
y° = 180° – 120°
y° = 60°

(OR)

y° + 120° = 1800 (linear pair of angles)
y° = 180°- 120°
∴ y° = 60°

ii) In the figure ΔRST,
x° = 80° (vertically opposite angles)
also ∠R + ∠S + ∠T = 180° (angle – sum property)
80° + 50°+ y°= 180°
130° + y° = 180°
y° = 180° – 130°
∴ y° = 50°

iii) m ΔMAN,
x° = ∠M + ∠A (exterior angle property)
x° = 50° + 60°
x° = 110°
Also x° + y° = 180°
110°+ y°= 180°
y° = 180° – 110°
y° = 70°

(OR) in ΔMAN,
∠M + ∠A + ∠N = 180° (angle – sum property )
50° + 60° + y° = 180°
110° + y° = 180°
y° = 180°- 110°
∴ y° = 70°

iv) In the figure ΔABC,
x° = 60° (vertically opposite angles)
∠A + ∠B + ∠ACB = 180° (angle – sum property)
y° + 30° + 60° = 180°
y° + 90° = 180°
y° = 180° – 90°
∴ y° = 90°

v) In the figure ΔEFG,
y° = 90° (vertically opposite angles)
Also in ΔEFG;
∠F + ∠E + ∠G = 180° (angle – sum property)
∴ x° + x° + y° = 180
2x° + 90° = 180°
2x° = 180°- 90°
2x° = 90°
x° = \(\frac{90^{\circ}}{2}\)
∴ x° = 45°

vi) In the figure ΔLET,
∠L = ∠T = ∠E = x° (vertically opposite angles)
Also in ΔLET
∠L + ∠E + ∠T = 180° (angle – sum property)
x° + x° + x° = 180°
3x° = 180°
x° = \(\frac{180^{\circ}}{3}\)
x° = 60°

Question 3.
Find the measure of the third angle of triangles whose two angles are given below:
(i) 38° , 102°
(ii) 116°, 30°
(iii) 40°, 80°
Solution:
(i) 38° , 102°
Let the third angle be x° then
38° + 102° + x° = 180° (angle – sum property)
140° + x° = 180°
x° = 180° – 140° = 40°

(ii) 116°, 30°
Let the third angle be x°
then 116° + 300 + x° = 180° (angle – sum property)
146° + x = 180°
x= 180°- 146° = 34°

(iii) 40°, 80°
Let the third angle be x° then
40C + 80° + x° = 180° (angle-sum property)
120°+x° = 180°
x° = 180 – 120° – 60°.

Question 4.
In a right-angled triangle, one acute angle is 30°. Find the other acute angle.
Solution:
Given triangle is right angled triangle.
Let the third angle be x°
then 90° +30° + x° = 180° (angle – sum property)
120° + x° – 180°
x° = 180° – 120° = 60°

Question 5.
State true or false for each of the following statements.
(i) A triangle can have two right angles.
(ii) A triangle can have two acute angles.
(iii) A triangle can have two obtuse angles.
(iv) Each angle of a triangle can be less than 60°.
Solution:
(i) A triangle can have two right angles. – False
(ii) A triangle can have two acute angles. – True
(iii) A triangle can have two obtuse angles. – False
(iv) Each angle of a triangle can be less than 60°. – False

Question 6.
The angles of a triangle are in the ratio 1 : 2 : 3. Find the angles.
Solution:
Given that ratio of the angle = 1: 2: 3
Sum of the terms of the ratio – 1 + 2 + 3 = 6
Sum of the angles of a triangle = 1800
∴ 1st angle = \(\frac { 1 }{ 6 }\) x 180° = 30° (angle-sum property)
∴ 2nd angle = \(\frac { 2 }{ 6 }\) x 180° = 60° (angle – sum property)
∴ 3rd angle = \(\frac { 3 }{ 6 }\) x 180° = 90° (angIe – sum property)

Question 7.
In the figure, \(\overline{\mathrm{DE}} \| \overline{\mathrm{BC}}\) , ∠A = 300 and ∠B = 50°. Find the values of x, y and z.
Solution:

y° = 50° (correspondIng angles)
x° = z° (corresponding angles)
Also in MBC;
∠A + ∠B +∠C = 180° (angIe – sum property)
30° + 50° + z° = 180°
80° + z° = 180°
z° = 180° – 80°
∴ z° = 100°
∴ x° = 100°; y° = 50°; z° = 100°

Question 8.
In the figure, ∠ABD = 3 ∠DAB and ∠BDC = 96°. Find ∠ABD.
Solution:

In the figure
∠ABD + ∠DAB = 96° (exterior angle property)
3∠DAB + ∠DAB = 96° (given)
4∠DAB = 96°
∠DAB = \(\frac{96^{\circ}}{4}\) = 24°
∠ABD = 3 x 24° = 72°

Question 9.
In ΔPQR ∠P= 2 ∠Q and 2 ∠R =3 ∠Q , calculate the angles of ΔPQR.
Solution:
In ΔPQR
∠P + ∠Q + ∠R = 180° also
∠P:∠Q:∠R = 2∠Q:∠Q: \(\frac { 3 }{ 2 }\)∠Q = 4 : 2 : 3
Sum of the term of the ratio = 4 + 2 + 3 = 9
∠P = \(\frac { 4 }{ 9 }\) x 180° = 80°
∠Q = \(\frac { 2 }{ 9 }\) x 180° = 40°
∠R = \(\frac { 3 }{ 9 }\) x 180° = 60°

Question 10.
If the angles of a triangle are in the ratio 1:4: 5, find the angles.
Solution:
Given that ratio of the angles = 1 : 4 : 5
Sum of the terms of the ratio = 1 + 4 + 5 = 10
Sum of the angles 180°
∴ 1st angIe = \(\frac { 1 }{ 10 }\) x 180° = 18°
2nd angle \(\frac { 4 }{ 10 }\) x 180° = 72°
3rd angle = \(\frac { 5 }{ 10 }\) x 180°= 90°

Question 11.
The acute angles of a right triangle are in the ratio 2 : 3. Find the angles of the triangle.
Solution:
Given that ratio of acute angles = 2 : 3
Sum of the terms of the ratio = 2 + 3 = 5
Sum of the acute angles = 90°
∴ 1st acute angle x 900 = 36°
2nd acute angle = x 900 = 540
∴ Angles of the triangle = 36°, 54° and 90°

Question 12.
In the figure, ∆PQR is right angled at Q, \(\overline{\mathrm{ML}} \| \overrightarrow{\mathrm{RQ}}\) and ∠LMR = 130°. Find ∠LPM, ∠PML and ∠PRQ.

Solution:
∠PRQ + ∠LMR = 180° (int, angles on the same side of the transversal)
∠PRQ + 130° = 18 0°
∴ ∠PRQ = 180° – 130°= 50°
In ∆PRQ,
∠P + ∠R +∠Q = 180° (angle – sum property)
∠LPM + 50° + 90° = 180°
∠LPM + 140° = 180°
∴ ∠LPM = 180° – 140° = 40°
Also ∠PML + ∠LMR = 180° (Linear pair of angles)
∴ ∠PML + 130° = 180°
∠PML = 180° – 130° = 50°
(or) ∠PML = ∠PRQ = 50° (corresponding angles)

Question 13.
In Figure ABCDE, find ∠1 +∠2 + ∠3 + ∠4 + ∠5.

Solution:
In ∆ABC, ∠BAC + ∠ACB + ∠ABC = 180° ………….(1)
In ∆ACD, ∠CAD + ∠ADC + ∠ACD = 180° ……………. (2)
In ∆ADE, ∠DAE + ∠ADE + ∠AED = 180° …………(3)
Adding (1), (2) and (3) we get
(∠BAC + ∠ACB + ∠ABC) + ∠CAD+ (∠ADC + ∠ACD + ∠DAE + ∠ADE + ∠AED = 180°+ 180° 180°
(∠BAC + ∠CAD + ∠DAE) + ∠ABC + (∠ACB + ∠ACD) + ∠ADC + ∠ADE) + ∠AED = 180° + 180° + 180°
Hene
∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 540°

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 2 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 2

Question 1.
In ΔABC, D is the midpoint of \(\overline{\mathrm{BC}}\)
(i) \(\overline{\mathrm{AD}}\) is the ___________________
(ii) \(\overline{\mathrm{AE}}\) is the ____________________

Solution:
(i) \(\overline{\mathrm{AD}}\) is the median
(ii) \(\overline{\mathrm{AE}}\) is the Altitude

Question 2.
Name the triangle in which the two altitudes of the triangle are two of its sides.
Solution:
In Right angled triangle, the sides containing the right
angle are two altitudes.
In ΔCAT, ∠A = 90° and CA; AT are altitudes.

Question 3.
Does a median always lie in the interior of the triangle?
Solution:
Yes, a median always lie in the interior of the triangle.

Question 4.
Does an altitude always lie in the interior of a triangle’?
Solution:
No, an altitude need not always lie In the interior of a triangle.

Question 5.
(i) Write the side opposite to vertex Y in ΔXYZ.
(ii) Write the angle opposite to side \(\overline{\mathrm{PQ}}\) in ΔPQR.
(iii) Write the vertex opposite to side \(\overline{\mathrm{AC}}\) in ΔABC.
Solution:
i) Side opposite to vertex Y = \(\overline{\mathrm{XZ}}\)
ii) Angle opposite to side \(\overline{\mathrm{PQ}}\) = ∠R
lii) Vertex opposite to side \(\overline{\mathrm{AC}}\) = B

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 1

Question 1.
Is it possible to have a triangle with the following sides?
(i) 3 cm, 4 cm and 5 cm.
(ii) 6 cm, 6 cm and 6 cm.
(iii) 4 cm, 4 cm and 8 cm.
(iv) 3 cm, 5 cm and 7 cm.
Solution:
i) 3 + 4 > 5 ; 4 + 5 > 3 ; 3 5 > 4, ∴ Yes, a triangle can be formed with these sides.
ii) ∴Yes. A triangle can be formed with these sides.
iii) 4 + 4 > 8
∴ With these sides, a triangle cannot he formed.
iv) 3 + 5 > 7; 5 + 7 > 3; 3 + 7 > 5
∴ Yes. A triangle can he formed with these sides.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 4 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 4

Question 1.
Find the reciprocal of each of the following fractions.
(i) \(\frac { 5 }{ 8 }\)
(ii) \(\frac { 8 }{ 7 }\)
(iii) \(\frac { 13 }{ 7 }\)
(iv) \(\frac { 3 }{ 4 }\)
Solution:
i) Reciprocal of \(\frac { 5 }{ 8 }\) = \(\frac { 8 }{ 5 }\)
ii) Reciprocal of \(\frac { 8 }{ 7 }\) = \(\frac { 7 }{ 8 }\)
iii) Reciprocal of \(\frac { 13 }{ 7 }\) = \(\frac { 7 }{ 13 }\)
iv) Reciprocal of \(\frac { 3 }{ 4 }\) = \(\frac { 4 }{ 3 }\)

Question 2.
Find
(i) 18 ÷ \(\frac { 3 }{ 4 }\)
(ii) 8 ÷ \(\frac { 7 }{ 3 }\)
(iii) 3 ÷ 2\(\frac { 1 }{ 3 }\)
(iv) 5 ÷ 3\(\frac { 4 }{ 7 }\)
Solution:

Question 3.
Find
(i) \(\frac { 2 }{ 5 }\) ÷ 3
(ii) \(\frac { 7 }{ 8 }\) ÷ 5
(iii) \(\frac { 4 }{ 9 }\) ÷ \(\frac { 4 }{ 5 }\)
Solution:

Question 4.
Deepak can paint \(\frac { 2 }{ 5 }\) of a house in one day. lfhe continues working at this rate, how many days will he take to paint the whole house?
Amount of work done by deepak in 1 day = \(\frac { 2 }{ 5 }\) th
Total work can be done in = 1 ÷ \(\frac { 2 }{ 5 }\) days
= \(\frac { 2 }{ 5 }\) = \(1 \times \frac{5}{2}=\frac{1 \times 5}{2}=\frac{5}{2}=2 \frac{1}{2}\) days.

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 2 Fractions, Decimals and Rational Numbers Ex 3 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 2nd Lesson Fractions, Decimals and Rational Numbers Exercise 3

Question 1.
Find each of the following products.
(i) \(\frac{5}{6} \times \frac{7}{11}\)
(ii) \(6 \times \frac{1}{5}\)
(iii) \(2 \frac{1}{3} \times 3 \frac{1}{5}\)
Solution:

Question 2.
Multiply and reduce to lowest form.
(i) \(\frac{2}{3} \times 5 \frac{1}{5}\)
(ii) \(\frac{2}{7} \times \frac{1}{3}\)
(iii) \(\frac{9}{3} \times \frac{5}{5}\)
Solution:

Question 3.
Which one is greater’?
(i) \(\frac{2}{5} \text { of } \frac{4}{7} \text { or } \frac{3}{4} \text { of } \frac{1}{2}\)
(ii) \(\frac{1}{2} \text { of } \frac{4}{7} \text { or } \frac{2}{3} \text { of } \frac{3}{7}\)
Solution:
(i) \(\frac{2}{5} \text { of } \frac{4}{7} \text { or } \frac{3}{4} \text { of } \frac{1}{2}\)

(ii) \(\frac{1}{2} \text { of } \frac{4}{7} \text { or } \frac{2}{3} \text { of } \frac{3}{7}\)

Question 4.
Rehana works 2\(\frac{1}{2}\) hours each day on her embroidery. She completes the work in 7 days.
How many hours did she take to complete her work?
Solution:
Number of hours Rehana worked each day = 2\(\frac{1}{2}\) hrs
Number of days Rehana worked on embroidery = 7 days
Total time worked = 2½ x 7 = \(\frac{5}{2}\) x 7 = \(\frac{5 \times 7}{2}=\frac{35}{2}\) = = 17\(\frac{1}{2}\) hrs.

Question 5.
A truck runs 8 km using 1 litre of petrol. How much distance will it cover using 10\(\frac{2}{3}\) litres of petrol?
Solution:
bistance covered wIth 1 litre of petrol 8 km
∴ Distance covered for 10\(\frac{2}{3}\) litres petrol = 10\(\frac{2}{3}\) x 8
= \(\frac{32}{3} \times 8=\frac{32 \times 8}{3}=\frac{256}{3}=85 \frac{1}{3}\)km.

Question 6.
Raja walks 1\(\frac{1}{2}\) meters in 1 second. How much distance will he walk in 15 minutes?
Solution:
Distance covered by Raja in 1 second = 1\(\frac{1}{2}\) m
Distance covered by Raja in 15 minutes 15 x 60 x 1\(\frac{1}{2}\)
= 900 x \(\frac{3}{2}\) = \(\frac{900\times 3}{21}\) =1350 m

Question 7.
Provide the number in the box to make the statement true.